What is the time complexity of Find the Index of the First Occurrence in a String?

The optimal solution has a time complexity of O(m*n).

What data structures are used to solve Find the Index of the First Occurrence in a String?

This problem uses Two Pointers, String, String Matching. Understanding these concepts is essential for solving this Easy problem efficiently.

What are the key insights for solving Find the Index of the First Occurrence in a String?

**Sliding Window Comparison:** The core idea is to simulate a 'window' of the `needle`'s length sliding across `haystack` from left to right. At each position `i`, we effectively create a temporary substring of `haystack` (from `i` to `i + len(needle) - 1`) and compare it against the entire `needle`. This systematic approach guarantees that the first possible match will be found. **Precise Loop Bounds:** The loop iterating through `haystack` needs careful bounds. It runs from `i = 0` up to `len(haystack) - len(needle)`. This upper bound ensures that `i + len(needle)` never exceeds `len(haystack)`, preventing attempts to access characters beyond the string's end and ensuring that the extracted substring is always of the correct length to match `needle`.

Which companies ask LeetCode 28. Find the Index of the First Occurrence in a String in interviews?

This problem is frequently asked at Accenture, Adobe, Amazon, Apple, Bloomberg and 14 more companies.

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Find the Index of the First Occurrence in a String - LeetCode 28 Solution

Q: How to solve Find the Index of the First Occurrence in a String?

The provided solution employs a straightforward, brute-force 'sliding window' approach. It systematically checks every possible starting position in `haystack` where `needle` *could* begin. The algorithm iterates through `haystack` using an index `i`, which represents a potential starting point for `needle`. For each `i`, it extracts a substring from `haystack` that is exactly the same length as `needle`, specifically `haystack[i : i + len(needle)]`. It then directly compares this extracted substring with `needle` itself. If a match is found (`haystack[i : i + len(needle)] == needle`), the current index `i` is immediately returned because we are specifically looking for the *first* occurrence. If the loop completes without finding any match, it implies that `needle` is not present in `haystack`, and the function returns `-1`. The initial check for an empty `needle` is a critical edge case, adhering to the common convention that an empty string is considered to occur at index 0 of any string (even an empty one).

Find the Index of the First Occurrence in a String - Complete Solution Guide

Find the Index of the First Occurrence in a String is LeetCode problem 28, a Easy level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

Given two strings needle and haystack , return the index of the first occurrence of needle in haystack , or -1 if needle is not part of haystack . Example 1: Input: haystack = "sadbutsad", needle = "sad" Output: 0 Explanation: "sad" occurs at index 0 and 6. The first occurrence is at index 0, so we return 0. Example 2: Input: haystack = "leetcode", needle = "leeto" Output: -1 Explanation: "leeto" did not occur in "leetcode", so we return -1. Constraints: 1 <= haystack.length, needle.length <= 10

Detailed Explanation

This problem, "Find the Index of the First Occurrence in a String," asks us to locate the starting position of a smaller string, `needle`, within a larger string, `haystack`. If `needle` appears multiple times, we're only interested in the very first one, meaning the one with the lowest starting index. If `needle` doesn't exist anywhere in `haystack`, our job is to return `-1`. The task is essentially a fundamental string search operation, a building block for many text processing functionalities.

Solution Approach

The provided solution employs a straightforward, brute-force 'sliding window' approach. It systematically checks every possible starting position in `haystack` where `needle` *could* begin. The algorithm iterates through `haystack` using an index `i`, which represents a potential starting point for `needle`. For each `i`, it extracts a substring from `haystack` that is exactly the same length as `needle`, specifically `haystack[i : i + len(needle)]`. It then directly compares this extracted substring with `needle` itself. If a match is found (`haystack[i : i + len(needle)] == needle`), the current index `i` is immediately returned because we are specifically looking for the *first* occurrence. If the loop completes without finding any match, it implies that `needle` is not present in `haystack`, and the function returns `-1`. The initial check for an empty `needle` is a critical edge case, adhering to the common convention that an empty string is considered to occur at index 0 of any string (even an empty one).

Step-by-Step Algorithm

Step 1: Check for edge cases: If the needle is empty, return 0. If the needle is longer than the haystack, return -1.
Step 2: Iterate through the haystack string using a loop. The loop runs from index 0 up to `len(haystack) - len(needle)`.
Step 3: In each iteration, extract a substring from the haystack of the same length as the needle, starting at the current index `i`.
Step 4: Compare the extracted substring with the needle. If they are equal, return the current index `i`.
Step 5: If the loop completes without finding a match, return -1.

Key Insights

**Sliding Window Comparison:** The core idea is to simulate a 'window' of the `needle`'s length sliding across `haystack` from left to right. At each position `i`, we effectively create a temporary substring of `haystack` (from `i` to `i + len(needle) - 1`) and compare it against the entire `needle`. This systematic approach guarantees that the first possible match will be found.
**Precise Loop Bounds:** The loop iterating through `haystack` needs careful bounds. It runs from `i = 0` up to `len(haystack) - len(needle)`. This upper bound ensures that `i + len(needle)` never exceeds `len(haystack)`, preventing attempts to access characters beyond the string's end and ensuring that the extracted substring is always of the correct length to match `needle`.
**Leveraging Built-in Substring Operations (Python's Slicing):** Python's slicing syntax (`haystack[i:i + len(needle)]`) is a remarkably concise and often optimized way to extract a substring. This built-in language feature abstracts away the manual character-by-character comparison within the window, making the code exceptionally readable and less prone to off-by-one errors than a manual nested loop for comparison.

Complexity Analysis

Time Complexity: O(m*n)

Space Complexity: O(1)

Topics

This problem involves: Two Pointers, String, String Matching.

Companies

Asked at: Accenture, Adobe, Amazon, Apple, Bloomberg, Expedia, Google, IBM, IIT Bombay, Infosys, Meta, Microsoft, Pocket Gems, Uber, Warnermedia, Yahoo, Yandex, Zoho, tcs.

Find the Index of the First Occurrence in a String | LeetCode 28

Easy

Two Pointers String String Matching

Time: O(m*n)Space: O(1)

Solution Code

Available in:

class Solution {
    public int strStr(String haystack, String needle) {
        int n = haystack.length();
        int m = needle.length();

        if (m == 0) {
            return 0;
        }

        for (int i = 0; i <= n - m; i++) {
            if (haystack.substring(i, i + m).equals(needle)) {
                return i;
            }
        }

        return -1;
    }
}

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Problem Info

Description

Problem 28: Find the Index of the First Occurrence in a String - Detailed explanation and solution approaches.

Topics

Two Pointers String String Matching

Companies

AccentureAdobeAmazonAppleBloombergExpediaGoogleIBM

Stats

Difficulty:Easy

Languages:4

Companies:19

Problem #:28

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

5 steps

1
Step 1: Check for edge cases: If the needle is empty, return 0. If the needle is longer than the haystack, return -1.
2
Step 2: Iterate through the haystack string using a loop. The loop runs from index 0 up to `len(haystack) - len(needle)`.
3
Step 3: In each iteration, extract a substring from the haystack of the same length as the needle, starting at the current index `i`.
4
Step 4: Compare the extracted substring with the needle. If they are equal, return the current index `i`.
5
Step 5: If the loop completes without finding a match, return -1.

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(m*n)

💾

Space Complexity

O(1)

💪 Constant space - Uses same memory regardless of input!

📝

Why?

The nested loops (implicit in the substring comparison) lead to O(m*n) time complexity. The algorithm uses a constant amount of extra space regardless of input size, making it O(1) space.

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

Knuth-Morris-Pratt (KMP) Algorithm

A more efficient string matching algorithm that preprocesses the needle to create a failure function. This function helps avoid redundant comparisons when a mismatch occurs.

Time: O(n + m)Space: O(m)

Trade-offs: KMP is significantly faster for larger needles and multiple searches, but adds the overhead of preprocessing the needle.

Rabin-Karp Algorithm

Uses hashing to compare substrings more efficiently, but can be affected by hash collisions.

Time: Average O(n+m), Worst Case O(mn)Space: O(1)

Trade-offs: Rabin-Karp offers potential speed advantages but depends on the quality of the hash function and can be slower in the worst case due to collisions

✨

Key Insights

1**Sliding Window Comparison:** The core idea is to simulate a 'window' of the `needle`'s length sliding across `haystack` from left to right. At each position `i`, we effectively create a temporary substring of `haystack` (from `i` to `i + len(needle) - 1`) and compare it against the entire `needle`. This systematic approach guarantees that the first possible match will be found.
2**Precise Loop Bounds:** The loop iterating through `haystack` needs careful bounds. It runs from `i = 0` up to `len(haystack) - len(needle)`. This upper bound ensures that `i + len(needle)` never exceeds `len(haystack)`, preventing attempts to access characters beyond the string's end and ensuring that the extracted substring is always of the correct length to match `needle`.
3**Leveraging Built-in Substring Operations (Python's Slicing):** Python's slicing syntax (`haystack[i:i + len(needle)]`) is a remarkably concise and often optimized way to extract a substring. This built-in language feature abstracts away the manual character-by-character comparison within the window, making the code exceptionally readable and less prone to off-by-one errors than a manual nested loop for comparison.

📚

Key Terms Explained

For Students

Two Pointers

Using two indices to traverse data from different positions simultaneously.

String

A sequence of characters. Like a word or sentence in programming!

Edge Cases

Edge case 1: Empty needle. The code correctly handles this by returning 0.
Edge case 2: Needle longer than haystack. The code handles this by returning -1 before the loop even starts.
Edge case 3: Needle not found in haystack. The code correctly returns -1 after iterating through the entire haystack.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Forgetting to handle the edge case of an empty needle. This should return 0.

✗Common mistake 2: Incorrectly calculating the loop bounds, leading to index out of bounds errors or missing potential matches.

✗Common mistake 3: Inefficient substring comparisons. Repeated substring creation can be optimized with careful index management in some languages.

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: Searching for specific patterns in large text files (e.g., log files, documents).

🏦Real-world scenario 2: Implementing search functionality in text editors or IDEs.

🎮Real-world scenario 3: Network security (intrusion detection systems looking for malicious patterns in network traffic).

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

The two-pointer technique is powerful for sorted arrays. The key insight is recognizing when elements can be processed from both ends simultaneously.

🎯 Interview Tip

Use this problem to demonstrate clean code practices. Variable naming and code organization matter.

⚠️ Watch Out

The most common mistake is not handling the case when pointers meet. Always consider the termination condition.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: June 29, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

Try solving this without using extra space first. It's a great exercise for understanding the underlying pattern.

@CodeNinjaJul 14

In a real interview, always clarify constraints first. Ask about input size and value ranges.

@TechLeadJul 7

❓

Frequently Asked Questions

When is the two-pointer technique most effective?

Two pointers work best on sorted arrays or when searching for pairs/triplets that satisfy a condition. Common patterns include: moving pointers inward from both ends (like in two-sum variants), fast/slow pointers for cycle detection, or left/right pointers for partitioning.

How do I handle string manipulation efficiently?

In languages with immutable strings (Java, Python, C#), use StringBuilder/list to avoid O(n) concatenation cost. Consider using character arrays for in-place modifications. For pattern matching, consider KMP or Rabin-Karp for O(n+m) complexity.

How should I approach easy-level algorithm problems?

Easy problems build foundational skills. Focus on: 1) Understanding the problem completely before coding, 2) Writing clean, readable code, 3) Handling all edge cases (empty input, single element, etc.), 4) Analyzing time and space complexity. Don't skip complexity analysis even for simple solutions.

How should I practice algorithm problems effectively?

Focus on understanding patterns rather than memorizing solutions. After solving a problem, review optimal solutions and understand the intuition. Group similar problems to recognize patterns. Practice explaining your approach out loud. Review problems after a few days to test retention.

What is the optimal approach for "Find the Index of the First Occurrence in a String"?

For this easy-level Two Pointers problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether string techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

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